Submission #1071575

#	Time	Username	Problem	Language	Result	Execution time	Memory
1071575		bleahbleah	Beech Tree (IOI23_beechtree)	C++17	51 / 100	618 ms	164016 KiB

beechtree

#include "beechtree.h"
#include <bits/stdc++.h>
#define all(x) (x).begin(),(x).end()
using namespace std;

using ll = long long;
using ld = long double;

//#define int ll
#define sz(x) ((int)(x).size())

using pii = pair<int,int>;
using tii = tuple<int,int,int>;

const int nmax = 3e5 + 5;

const int mod = 998244853;
struct Mint {
  int val;
  Mint(ll x = 0): val((x % mod + mod) % mod) {;}
  Mint operator +(const Mint& x) const { return Mint(val + x.val); }
  Mint operator -(const Mint& x) const { return Mint(val - x.val); }
  Mint operator *(const Mint& x) const { return Mint((ll)val * x.val); }
  Mint operator +=(const Mint& x) { return *this = Mint(val + x.val); }
  Mint operator -=(const Mint& x) { return *this = Mint(val - x.val); }
  Mint operator *=(const Mint& x) { return *this = Mint((ll)val * x.val); }
  Mint operator ^(const int& _b) const {
    Mint accum = 1, a = *this;
    int b = _b;
    while(b) {
      accum = (b & 1? accum * a : accum);
      a *= a;
      b >>= 1;
    }
    return accum;
  }
  Mint operator /(const Mint& x) { return Mint((ll)val * (x ^ (mod - 2)).val); }
  Mint operator /=(const Mint& x) { return *this = Mint((ll)val * (x ^ (mod - 2)).val); }
};

Mint p[2][nmax];

#define hash bjsefdjhsdsfhoi
struct hash {
   Mint v[2];
   int len;
   hash(Mint a, Mint b, int c) { v[0] = a; v[1] = b; len = c; }
   hash(Mint a) { v[0] = a; v[1] = a; len = 1; }
   hash() { v[0] = 0; v[1] = 0; len = 0; }
   hash operator +(const hash& x) const {
      return hash(v[0] * p[0][x.len] + x.v[0], v[1] * p[1][x.len] + x.v[1], len + x.len);
   }
   hash operator -(const hash& x) const {
      return hash(v[0] - p[0][len - x.len] * x.v[0], v[1] - p[1][len - x.len] * x.v[1], len - x.len); 
   }
   hash operator += (const hash& x) { return *this = *this + x; }
   hash operator -= (const hash& x) { return *this = *this - x; }
   bool operator !=(const hash& x) const {
      return v[0].val != x.v[0].val || v[1].val != x.v[1].val || len != x.len;
   }
   ll operator()() const { return (ll)v[0].val * mod + v[1].val; }
   bool operator < (const hash& x) const { return (*this)() < x(); }
   bool operator == (const hash& x) const { return (*this)() == x(); }
};



vector<pii> g[nmax];
vector<pii> invg[nmax];
vector<int> P, C;

bool isanc(unordered_set<int>& A, unordered_set<int>& B) {
   if(sz(A) < sz(B)) return 0;
   for(auto &x : B)
      if(!A.count(x)) return 0;
   return 1;
}

vector<int> sol;
int area[nmax], pin[nmax], pout[nmax], inp;
hash subarb[nmax];

hash eulerH[nmax], dH[nmax];
pii euler[nmax];
int poz;


void init(int node) {
   sort(all(g[node]), [&](auto a, auto b) { return a.second < b.second; });
   euler[++poz] = pii{node, 0};
   eulerH[poz] = eulerH[poz - 1] + hash(2 * C[node] + 0);
   pin[node] = poz + 1;
   
   area[node] = 1;
   for(auto [x, c] : g[node])
      dH[x] = dH[node] + hash(c), init(x), area[node] += area[x];
      
   pout[node] = poz;
   euler[++poz] = pii{node, 1};
   eulerH[poz] = eulerH[poz - 1] + hash(2 * C[node] + 1);
}

vector<int> difference(int node, int other) {
   vector<int> ass_list;
   auto getL = [&](int l, int r) {
      return eulerH[r + pin[node]] - eulerH[l + pin[node] - 1];
   };
   auto getR = [&](int l, int r) {
      return eulerH[r + pin[other]] - eulerH[l + pin[other] - 1];
   };
   
   auto visit = [&](auto&& self, int nod, int& r) -> void {
      //cerr << nod << '\n';
      r++;
      ass_list.emplace_back(nod);
      for(auto [x, c] : g[nod]) self(self, x, r);
      r++;
   };
   
   int l0 = 0, r0 = 0, END0 = pout[node] - pin[node] + 1, l1 = 0, r1 = 0, END1 = pout[other] - pin[other] + 1;
   
   //cerr << node << ", " << other << '\n';
   //for(int i = 0; i < END0; i++) cerr << getL(i, i).v[0].val << ' '; cerr << '\n';
   //for(int i = 0; i < END1; i++) cerr << getR(i, i).v[0].val << ' '; cerr << '\n';
   
   while(l0 < END0 && l1 < END1) {
      r0 = l0 - 1, r1 = l1 - 1;
      for(int lim = 1 << 18; lim > 0; lim >>= 1) {
         if(r0 + lim < END0 && r1 + lim < END1 && getL(l0, r0 + lim)() == getR(l1, r1 + lim)()) r0 += lim, r1 += lim;
      }
      if(r0 == END0 - 1 || r1 == END1 - 1);
      else {
         auto [nxt0, t0] = euler[pin[node] + r0 + 1];
         auto [nxt1, t1] = euler[pin[other] + r1 + 1];
         //cerr << nxt0 << ' ' << t0 << '\t' << nxt1 << ' ' << t1 << '\t' << r0 << ", " << r1 << '\n';
         if(t1 == 0 && C[nxt1] < C[nxt0]) return {-1};
         if(t0 == 1) return {-1};
         visit(visit, nxt0, r0);
      }
      l0 = r0 + 1;
      l1 = r1 + 1;
   }
   //cerr << "<\\>\n";
   if(l0 == END0 && l1 < END1) return {-1};
   
   
   while(l0 < END0) {
      auto [nxt0, t0] = euler[pin[node] + l0];
      visit(visit, nxt0, l0);
   }
   //cerr << "<\\>\n";
   
   return ass_list;
}

#define erset set

struct PartDiff {
   map<int, erset<hash>> onlevel;
   map<hash, int> inverse;
   
   bool feasable = 1;
   
   void add(int dim, int prevdim, erset<hash> ths) {
      if(!feasable) return;
      if(onlevel.count(dim)) {
         for(auto x : ths) {
            if(inverse.count(x) == 0) { feasable = 0; return; }
            if(inverse[x] > dim) { feasable = 0; return; }
         }
      }
      else {
         auto it = onlevel.upper_bound(dim);
         int bigger = -1;
         
         if(it == onlevel.end()); else bigger = it -> first;
         
         
         for(auto x : ths) {
            if(inverse.count(x) == 0) { if(bigger == -1) { ; } else { feasable = 0; return; } }
            else {
               if(inverse[x] > dim) { 
                  if(inverse[x] == bigger) { onlevel[bigger].erase(x); } // o sa il pun, dar n-are ce cauta aici
                  else { feasable = 0; return; } // prost
               }
               else { assert(inverse[x] < dim); 
                  if(inverse[x] > prevdim) { continue; } // nu vreau sa il pun
                  else { feasable = 0; return; } // prost
               }
            }
            onlevel[dim].emplace(x);
            inverse[x] = dim;
         }
      }
   }
   void add(PartDiff& x) {
      if(!feasable) return;
      if(!x.feasable) { feasable = 0; return; }
      if(sz(x.inverse) > sz(inverse)) swap(x, *this); 
      int prv = -1;
      for(auto [h, v] : x.onlevel) {
         add(h, prv, v);
         prv = h;
      }
      x.onlevel.clear();
      x.inverse.clear();
      return;
   }
};


PartDiff dfs(int node) {
   PartDiff mine;
   int heavyson = -1;
   if(node == 0)
      sort(all(g[node]), [&](auto a, auto b) { return sz(g[a.first]) < sz(g[b.first]); });
   for(auto &[x, c] : g[node]) {
      auto T = dfs(x);
      //cout << x << " pula \n" << T.feasable << '\n';
      mine.add(T);  // asta teoretic e useless ca din superimpozabilitate toate vor fi deja acolo, e mai mult un fel de check :clown:
                    // oare checkul simplu poate fi facut mai putin cretin thinking thinking
      //cout << x << " pula \n" << T.feasable << '\n';
      if(heavyson == -1 || area[x] > area[heavyson]) heavyson = x;
      //cout << node << ' ' << x << '\t' << mine.feasable << ' ' << T.feasable << '\n';
      //if(node == 0) {
         //for(auto [t, c] : g[x]) cerr << c << ' ';
         //cerr << '\n';
      //}
   }
   if(sz(g[node]) != sz(invg[node])) {  mine.feasable = sol[node] = 0; return mine; }
   if(mine.feasable == 0) { sol[node] = 0; return mine; }
   
   if(heavyson == -1) {
      erset<hash> pl; pl.emplace(hash());
      mine.add(1, -1, pl);
   }
   else {
      auto sons = difference(node, heavyson);
      if(sz(sons) && sons[0] == -1) { mine.feasable = sol[node] = 0; return mine; }
      
      erset<hash> pl;
      for(auto x : sons)
         pl.emplace(dH[x] - dH[node]);
      mine.add(area[node], area[heavyson], pl);
   }
   sol[node] = mine.feasable;
   return mine;
}

std::vector<int> beechtree(int N, int M, std::vector<int> P_, std::vector<int> C_) {
   mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
   p[0][0] = p[1][0] = 1;
   p[0][1] = rng() % (mod - 1000) + 503;
   #ifdef DLOCAL
      p[0][1] = 10;
   #endif
   p[1][1] = rng() % (mod - 1200) + 505;
   for(int i = 2; i < nmax; i++)
      p[0][i] = p[0][i - 1] * p[0][1],
      p[1][i] = p[1][i - 1] * p[1][1];
      
   P = P_;
   C = C_;
   for(int i = 1; i < N; i++) {
      g[P[i]].emplace_back(i, C[i]);
      invg[P[i]].emplace_back(C[i], -1);
   }
   for(int i = 0; i < N; i++) sort(all(invg[i])), invg[i].erase(unique(all(invg[i])), end(invg[i]));
   sol.assign(N, 0);
   init(0);
   dfs(0);
   
   
   return sol;
}

/**
      Töte es durch genaue Untersuchung\Töte es kann es nur noch schlimmer machen\Es lässt es irgendwie atmen
--
*/ 

Compilation message (stderr)

beechtree.cpp: In function 'PartDiff dfs(int)':
beechtree.cpp:230:67: warning: suggest parentheses around assignment used as truth value [-Wparentheses]
  230 |    if(sz(g[node]) != sz(invg[node])) {  mine.feasable = sol[node] = 0; return mine; }
beechtree.cpp:239:65: warning: suggest parentheses around assignment used as truth value [-Wparentheses]
  239 |       if(sz(sons) && sons[0] == -1) { mine.feasable = sol[node] = 0; return mine; }

• Subtask 0: 0 / 0
• Subtask 1: 9 / 9
• Subtask 2: 5 / 5
• Subtask 3: 9 / 9
• Subtask 4: 0 / 8
• Subtask 5: 14 / 14
• Subtask 6: 14 / 14
• Subtask 7: 0 / 12
• Subtask 8: 0 / 17
• Subtask 9: 0 / 12